# Diode Transistor

Topics: P-n junction, Transistor, Diode Pages: 6 (2054 words) Published: March 14, 2013
The single-transistor inverter circuit illustrated earlier is actually too crude to be of practical use as a gate. Real inverter circuits contain more than one transistor to maximize voltage gain (so as to ensure that the final output transistor is either in full cutoff or full saturation), and other components designed to reduce the chance of accidental damage. Shown here is a schematic diagram for a real inverter circuit, complete with all necessary components for efficient and reliable operation:

This circuit is composed exclusively of resistors and bipolar transistors. Bear in mind that other circuit designs are capable of performing the NOT gate function, including designs substituting field-effect transistors for bipolar (discussed later in this chapter). Let's analyze this circuit for the condition where the input is "high," or in a binary "1" state. We can simulate this by showing the input terminal connected to Vcc through a switch:

In this case, diode D1 will be reverse-biased, and therefore not conduct any current. In fact, the only purpose for having D1 in the circuit is to prevent transistor damage in the case of a negative voltage being impressed on the input (a voltage that is negative, rather than positive, with respect to ground). With no voltage between the base and emitter of transistor Q1, we would expect no current through it, either. However, as strange as it may seem, transistor Q1 is not being used as is customary for a transistor. In reality, Q1 is being used in this circuit as nothing more than a back-to-back pair of diodes. The following schematic shows the real function of Q1:

The purpose of these diodes is to "steer" current to or away from the base of transistor Q2, depending on the logic level of the input. Exactly how these two diodes are able to "steer" current isn't exactly obvious at first inspection, so a short example may be necessary for understanding. Suppose we had the following diode/resistor circuit, representing the base-emitter junctions of transistors Q2 and Q4 as single diodes, stripping away all other portions of the circuit so that we can concentrate on the current "steered" through the two back-to-back diodes:

With the input switch in the "up" position (connected to Vcc), it should be obvious that there will be no current through the left steering diode of Q1, because there isn't any voltage in the switch-diode-R1-switch loop to motivate electrons to flow. However, there will be current through the right steering diode of Q1, as well as through Q2's base-emitter diode junction and Q4's base-emitter diode junction:

This tells us that in the real gate circuit, transistors Q2 and Q4 will have base current, which will turn them on to conduct collector current. The total voltage dropped between the base of Q1 (the node joining the two back-to-back steering diodes) and ground will be about 2.1 volts, equal to the combined voltage drops of three PN junctions: the right steering diode, Q2's base-emitter diode, and Q4's base-emitter diode. Now, let's move the input switch to the "down" position and see what happens:

If we were to measure current in this circuit, we would find that all of the current goes through the left steering diode of Q1 and none of it through the right diode. Why is this? It still appears as though there is a complete path for current through Q4's diode, Q2's diode, the right diode of the pair, and R1, so why will there be no current through that path? Remember that PN junction diodes are very nonlinear devices: they do not even begin to conduct current until the forward voltage applied across them reaches a certain minimum quantity, approximately 0.7 volts for silicon and 0.3 volts for germanium. And then when they begin to conduct current, they will not drop substantially more than 0.7 volts. When the switch in this circuit is in the "down" position, the left diode of the steering diode pair is fully conducting, and so it drops about...